Please refer to FIG. 1, it is a schematic circuit diagram of a conventional current source inverter 1, which has a current source output I_o, and can be electrically connected to a utility power (Utility) in parallel. In FIG. 1, the current source inverter 1 includes a buck converter 11 and a DC/AC converter 12. The buck converter 11 includes an input capacitor C_in, a buck converter switch S_buck, a free-wheeling diode D_buck and a output inductor L_o receiving a DC input voltage V_in and generating an output inductor current I_Lo. The DC/AC converter 12 includes a full bridge switching circuit 121, which includes four unidirectional switches and an output capacitor C_o. The four unidirectional switches are D1+Q1, D2+Q2, D3+Q3 and D4+Q4. In which, D1, D2, D3 and D4 are rectifying diodes, and Q1, Q2, Q3 and Q4 are power transistors, e.g., MOSFETs, IGBTs and bipolar transistors. An electrical power stored in the output inductor L_o is discharged to the output capacitor C_o when the output capacitor C_o is separated from the utility power (Utility), which will cause an AC output voltage V_o of the current source inverter 1 to be higher than the rated AC output voltage, and transistors having relatively higher tolerated voltage are required to avoid the damage of the power transistors Q1-Q4.
Referring to FIG. 1, an important function of the buck converter 11 is to cause a power factor of the current source inverter 1 to be one, which is resulting from that the output inductor current I_Lo would have one of an increase and a decrease following a variation of the waveform of the AC output voltage V_o of the current source inverter 1, and have a rectified sinusoidal waveform (see I_Lo as shown in FIG. 1) achieved via changing the duty cycle of the buck converter switch S_buck. The DC/AC converter 12 is employed to convert the rectified sinusoidal waveform of the output inductor current I_Lo into an AC sinusoidal waveform of an AC output current I_o (see FIG. 1).
Since the current flowing through the output inductor L_o, I_Lo, would vary according to the waveform of the AC output voltage V_o and have the rectified sinusoidal waveform when the current source inverter 1 is under operation, and generally speaking, the output inductor L_o has a relatively larger inductance, the buck converter 11 of FIG. 1 is equivalent to a current source 11 as shown in FIG. 2(a), which shows an equivalent circuit diagram of the conventional current source inverter 1 of FIG. 1 and a discharging route of the output inductor L_o when the utility power (Utility) is disconnected with the current source inverter 1, while the AC output voltage V_o is at a positive half-cycle.
Please refer to FIG. 2(a), the transistors Q1 and Q4 are conductive, the output inductor current I_Lo flows through the diodes D1 and D4, and the transistors Q1 and Q4 feedback an electrical power to the utility power (Utility) when the utility power (Utility) is connected with the current source inverter 1, while the AC output voltage is at the positive half-cycle. If the utility power (Utility) is disconnected with the current source inverter 1 at this moment (a socket 123 connected to and providing the utility power is separated from a plug 122), an electrical power stored in the output inductor L_o is discharged to the output capacitor C_o such that the AC output voltage V_o and the voltage V_1 across the transistor Q2 and the diode D2 are both increased. Similarly, the transistors Q2 and Q3 are conductive, the output inductor current I_Lo flows through the diodes D2 and D3, and the transistors Q2 and Q3 feedback an electrical power to the utility power (Utility) when the utility power (Utility) is connected with the current source inverter 1, while the AC output voltage V_o is at a negative half-cycle. If the utility power (Utility) is disconnected with the current source inverter 1 at this moment (the plug 122 is separated from the socket 123), the electrical power stored in the output inductor L_o is discharged to the output capacitor C_o such that the AC output voltage V_o and the voltage V_2 across the transistor Q4 and the diode D4 are both increased as shown in FIG. 2(b), which shows the equivalent circuit diagram of the conventional current source inverter 1 of FIG. 1 and a discharging route of the output inductor L_o when the utility power (Utility) is disconnected with the current source inverter 1, and the AC output voltage V_o is at the negative half-cycle. Due to the relatively higher voltages of V_1, V_2 and V_o, transistors having relatively higher tolerated voltage are required to avoid the damage of the power transistors Q1-Q4, and the manufacturing costs of which are increased, thus an improvement is needed.
Referring to FIG. 3(a), it is a schematic circuit diagram of a conventional current source inverter 2, which includes the buck converter 11, the DC/AC converter 12 having the full bridge switching circuit 121 (as shown in FIGS. 1, and 2(a)-2(b)) and an energy clamp circuit 21 having the first to the fourth diodes D_A, D_B, D_C and D_D. Due to that the DC input voltage V_in is larger than the peak value of the AC output voltage V_o when the current source inverter 2 is under a normal operation, the four diodes D_A, D_B, D_C and D_D are turned off. When the utility power (Utility) is disconnected with the current source inverter 2, the electrical power stored in the output inductor L_o would be feedbacked to the output capacitor C_o and the input capacitor C_in via one of the diodes D_A and D_D, and the diodes D_B and D_C. That is to say, the output inductor L_o discharges the stored electrical power to the output capacitor C_o and the input capacitor C_in when the utility power (Utility) is disconnected with the current source inverter 2. When the AC output voltage V_o is increased to a value such that the voltage across the output capacitor C_o is a little bit larger than the DC input voltage V_in, the diodes D_A, D_B, D_C and D_D are turned on, and the electrical power stored in the output inductor L_o is feedbacked to the terminals of the DC input power supply V_in (i.e. C_in). Thus, the AC output voltage V_o would be clamped at a level that is a little bit larger than that of the DC input voltage V_in such that voltage stresses of the transistors Q1 to Q4 could be decreased dramatically.
In FIG. 3(b), the transistors Q1 and Q4 are conductive, the output inductor current I_Lo flows through the diodes D1 and D4, and the transistors Q1 and Q4 feedback the electrical power stored in the output inductor L_o to the utility power (Utility) when the AC output voltage V_o of the utility power (Utility) is at the positive half-cycle. If the AC output voltage V_o of the utility power (Utility) is disconnected with the current source inverter 2 (the plug 122 is separated from the socket 123) at this moment, the electrical power stored in the output inductor L_o would be discharged to the output capacitor C_o and the input capacitor C_in via a positive half-cycle discharging route, i.e. the current route as shown in FIG. 3(b) (the output inductor L_o→the diode D1→the transistor Q1→the diode D_A→the input capacitor C_in→the diode D_D→the diode D4→the transistor Q4→the free-wheeling diode D_buck→the output inductor L_o).
Furthermore, referring to FIG. 3(c), the transistors Q2 and Q3 are conductive, the output inductor current I_Lo flows through the diodes D2 and D3, and the transistors Q2 and Q3 feedback the electrical power stored in the output inductor L_o to the utility power (Utility) when the AC output voltage V_o of the utility power (Utility) is at the negative half-cycle. If the AC output voltage V_o of the utility power (Utility) is disconnected with the current source inverter 2 (the plug 122 is separated from the socket 123) at this moment, the electrical power stored in the output inductor L_o would be discharged to the output capacitor C_o and the input capacitor C_in via a negative half-cycle discharging route, i.e. the current route as shown in FIG. 3(c) (the output inductor L_o→the diode D3→the transistor Q3→the diode D_B→the input capacitor C_in→the diode D_C→the diode D2→the transistor Q2→the free-wheeling diode D_buck→the output inductor L_o).
If the utility power (Utility) is connected to the current source inverter 2 of FIG. 3(a) when the DC input power supply V_in is not built up yet, a relatively large inrush current would be generated and flows through one of the diodes D_A and D_B, and the diodes D_C and D_D to the input capacitor C_in, the inrush current might burn down the fuse at the utility power (Utility) side, and would easily cause the interference of harmonic waves. The inrush current is generated flows through a positive half-cycle route (utility power→the first terminal of output capacitor C_o→diode D_A→input capacitor C_in→diode D_D→the second terminal of output capacitor C_o→utility power) as shown in FIG. 4(a) when the utility power (Utility) is connected with the current source inverter 2, the DC input voltage V_in is at one of the states being zero and being less than a peak value of the AC output voltage V_o of the utility power (Utility), and the AC output voltage V_o is at the positive half-cycle.
By the same token, the inrush current is generated and flows through a negative half-cycle route (utility power→the second terminal of output capacitor C_o→diode D_B→input capacitor C_in→diode D_C→the first terminal of output capacitor C_o→utility power) as shown in FIG. 4(b) when the utility power (Utility) is connected with the current source inverter 2, the DC input voltage V_in is at one of the states being zero and being less than a peak value of the AC output voltage V_o of the utility power (Utility), while the AC output voltage V_o is at the negative half-cycle.
Keeping the drawbacks of the prior arts in mind, and employing experiments and research full-heartily and persistently, the applicants finally conceived the current source inverter with the energy clamp circuit and the controlling method thereof having the relatively better effectiveness.